He Phuong Trinh Vi Phan Cap 1

8/8/2019 He Phuong Trinh Vi Phan Cap 1

1/26

H PHNG TRNH VI PHN CP 1


2/26

NH NGHA

F1(t,x1,x2,, xn, x1,x2,,xn) = 0.

Fn(t,x1,x2,, xn, x1,x2,,xn) = 0

H tng qut

x1 = f1(t,x1,x2,, xn).

xn = fn(t,x1,x2,, xn)

H chnh tc

t : bin

x1, x2 , , xn : n hm


3/26

BI TON CAUCHY

x1 = f1(t,x1,x2,, xn)

xn = fn(t,x1,x2,, xn)

Tm nghim h

Tha iu kin

x1(t0) = E1

..

xn(t0) = EnH n ptvp cp 1 tng ng 1 ptvp cp n nnh nghim c n hng s t do.


4/26

PHNG PHP KH

' '( ) 2

' '( ) 3

! !

! !

t

t

x x t y e

y y t x y e

B1: xy dng mt ptvp cp n theo 1 hm chn trc.

B2: gii ptvp cp n va tm c v rt v h vi (n 1)hm

Vd:

(1)

(2)

' 3 ' 2 3 '

' 2 ' 2

t t t

t t

y x y e y y e y e

x y e x y e

dd dd! !

! !

(3)


5/26

(3) " 3 ' 2 2 ! ty y y e Tt cp 2 h s hng

21 2 2 !

t t tt

(2) ' 3 ! tx y y e

21 22 (4 3)

t t tt!

21 2

21 2

2

2( 1) 3( 2 )

!

t t

t t t t tt t

21 2

21 2

2 (4 3)

2

!

!

t t t

t t t

t

t


6/26

Cch kh cho h 2 pt (tuyn tnh)

1.Ly o hm pt (1) theo t c (3)

2.Thay y t pt (2) vo (3) c (4)

3.Rt y t (1) thay vo (4)

4.Pt kt qu l pt cp 2 theo n hm x v bin t

1 1 1

2 2 2

( )

( )

x a x b y f t

y a x b y f t

d!

d!

(1)

(2)

Nu xut pht t pt (2), ta c pt cp 2 theo y


7/26

H PTVP TUYN TNH CP 1 H S HNG

1( )

( )

M

n

x t

x t

1( )

( )

d d

M

n

x t

x t

1( )

( )

M

n

f t

f t

X(t) = AX(t) + F(t)

11 1

1

: ma tran vuong cap n

!

L

L L L

L

n

n nn

a a

A

a a

(H n hm )


8/26

' '( ) 21 /

' '( ) 3

t

t

x x t y e

y y t x y e

! !

! !

( )( )( )

x ttt

!

0 2

1 3A !

( )

t

t

eF t

e

!

V d


9/26

2

sin1 1 2

( ) 2 4 1 ( ) ,

03 2 ln

t

t t

t t t

t

!

( )

( ) ( )

( )

x t

t t

z t

!

2

2 sin

2 / 2 4

3 2 lnt

x x y z t t

y x y z t

z y z t

!

!

!


10/26

PP TR RING GII H KHNG THUN NHT

X = AX + F(t)

A cho ha c

( P: P-1AP = D (cho) ) X = PDP-1X + F(t)

P-1X = DP-1X + P-1F(t)

t Y = P-1X:

Y = DY + G(t)

1 1 1 1

2 2 2 2

0 0 ( )

0 0 ( )

..... ..................... .... ...

0 0 ( )n n n n

y y g t

y y g t

y y g t

P

P

P

d

d ! d- - - -

K

K

K


11/26

1 1 1 1

2 2 2 2

0 0 ( )

0 0 ( )..... ..................... .... ...

0 0 ( )n n n n

y y g t

y y g t

y y g t

P

P

P

d

d ! d- - - -

K

K

K

1 11 1

2 22 2

( )

( )

..................................

(

( ) ( )

( ) ( )

( )) ( )n nn n

g t

g t

y t y t

y t y t

y tt t gy

P

P

P

d

d

! !

!d

H n ptvp tuyn tnh

cp 1

X = PY

gii

Y


12/26

1 2

2 1 2

2(1)

3

t

t

x x e

x x x e

d!

d! 0 2

,1 3

A

!

2 1

,1 1P

!

1Y P X!

1 1 1

,1 2P

!

( )

t

t

eF t

e

!

Cho ha A

1 0

,0 2D

!

1 11

2 2

y xP

y x

!


13/26

1(1) ( )Y DY P F td !

1 1

2 2

2

2 3

t

t

y y e

y y e

d!

d!

1 1 1 2

( ) 1 2 3

t t

t t

e e

P F t e e

! !

1 1

2 1

1 0 2

0 23

t

t

y y e

y ye

d ! d

1 1

22 2

2

3

t t

t t

y te C e

y e C e

!

!


14/26

1

2

2

22 1

1 13

t t

t t

te C e

e C e

!

21 1 2

22 1 2

( ) 2 4 3

( ) 2 3

t t t t

t t t t

x t C e C e te e

x t C e C e te e

!

!

Vy nghim h cho l:

1 1

2

2 2

2

3

t t

t t

y te C e

y e C e

!

!

21 2

2

1 2

2 4 3

2 3

t t t t

t t t t

C e C e te e

C e C e te e

!

X PY!


15/26

PPTR RING TM NGHIM H THUN NHT

1 1 1

2 2 2

( ) ( )( ) ( )

........................

( ) ( )

PP

P

! !

!n n n

y t y ty t y t

y t y t

X(t) = AX(t) Y = DY1 1 1

2 2 2

0 0

' 0 0

..... ..................... ....

' 0 0

P

P

P

! - - -

K

K

Kn n n

y y

y y

y y


16/26

1

2

1 1

2 2

....................

n

t

t

tn n

y t c e

y t c e

y t c e

P

P

P

!

!

!

1

k

nt

k

k

kX PY c e PP

!

! ! (Pk l ct th k ca P)

_ a, 1,.., :henghiem ltt cua hethuan nhat

k tk kX e P k n

P! !

1 1 1

2 2 2

'( ) ( )

'( ) ( )

........................

'( ) ( )

P

P

P

!

! !

n n n

y t y t

y t y t

y t y t


17/26

nh l: H X = AX(t), ma trn A c n gi tr ring

thc P1, P2 Pn (k c tr ring bi), v n vector

ring P1, P2 , , Pn c lp tuyn tnh

Nghim tng qut ca pt thun nht:

? A1 21

, , , P!

! ! K kn

T tn k kk

X t x t x t x t c e P


18/26

Vd:1 1 2 3

2 1 2 3

3 1 2 3

2

22 2 4

x x x x

x x x xx x x x

d!

d! d!

A

2

1 1 2

1 1 2 (6 ) 0

2 2 4

A IP

P P P P

P

! ! !

1

2

0

6

P

P

!

!

1 1 2

1 1 22 2 4

X X

d !


19/26

1( ) 0A I PP !1

2

3

1 1 2

1 1 2 0

2 2 4

p

p

p

!

Chn vector ring: 1 2

1 2

1 , 0

0 1

P P

! !

1

2 2

3

5 1 2

( ) 0 1 5 2 02 2 2

p

A I P pp

P

! ! Chn VTR: 3

1

1

2

P

!


20/26

1 11 1 2 2 ,,

t tX e P X e P

P P! !

3

1k k

k

X C X!

!

61 2 3

1 62 1 3

63 2 3

2

2

t

t

t

C C C ex

x C C e

x C C e

!

2 63 3 2

t tX e P e P

P! !

0 61 3

0

2

1 2

1 0

0 1

1

1

2

t t tC eC e C e

!


21/26

Cu trc nghim h tt khng thun nht

X = X0 + Xr

X0 : nghim tng qut h pt thunnht

X(t) = AX(t)

Xr: nghim ring h pt khng thunnht

Cu trc nghim tng qut ca h thun nht

X0 = C1X1 + C2X2 + + CnXn

{ Xk , k = 1, ..,n }: h nghim c lp tuyn tnh ca (1)

(1)


22/26

PP bin thin hng s tm Xr

Xr= C1(t)X1 + + Cn(t)Xn

C1(t)X1 + + Cn(t)Xn = F(t)Ci tm t h pt:


23/26

1 2

2 1 2

2(1)

3

t

tx x e

x x x e

d! d!

0 2 ,1 3

A !

( )

t

t

eF t

e

! H thun nht:

V d

1 2

2 1 2

2(2)

3

x x

x x x

d!

d!

1 1

21, ,

1PP

! !

Tr ring v VTR ca A:

1 2

12, ,

1PP

! !


24/26

Cc nghim ltt ca h thun nht

21 22 1,

1 1t tX e X e ! !

Nghim tng qut ca h thun nht

20 1 1 2 2 1 2

2 1

1 1

t tX C X C X C e C e

! !

Tm Xrbng pp bin thin hng s:

Trong X0 xem C1 v C2 l cc hm c theo t


25/26

Tm C1 v C2 t h:

C1(t)X1 + + Cn(t)Xn = F(t)

21 2

2 1

1 1

tt t

t

eC e C e

e

d d !

2

1 2

21 2

2t t t

t t t

C e C e e

C e C e e

d d !

d d !

1

2

2

3 t

C

C e

d!

d!

Chn:1

2

( ) 2

( ) 3t

C t t

C t e

!

!


26/26

20 1 1 2 2 1 2

2 1

1 1

t tX C X C X C e C e

! !

1

2

( ) 2

( ) 3t

C t t

C t e!

!

1 1 2 2( ) ( )rX C t X C t X !

22 1

2 31 1

t t t

te e e

! 4 3

2 3

t t

t t

te e

te e

!

0 rX X X! Nghim tng qut:

He Phuong Trinh Vi Phan Cap 1

Documents

Transcript of He Phuong Trinh Vi Phan Cap 1